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A thin uniform rod has a length of 0.550 m and is rotating in a circle on a frictionless table. The axis of rotation is perpendicular to the length of the rod at one end and is stationary. The rod has an angular velocity of 0.4 rad/s and a moment of inertia about the axis of 3.40×10−3 kg⋅m2. A bug initially standing on the rod at the axis of rotation decides to crawl out to the other end of the rod. When the bug has reached the end of the rod and sits there, its tangential speed is 0.202 m/s. The bug can be treated as a point mass.

a) What is the mass of the rod?
b) What is the mass of the bug?

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2 answers

  1. typo in 'Final I' -> should be .55^2 not .55^3

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  2. angular momentum = L = I omega
    Will not change because no torque.
    Original I = 3.40×10−3 kg⋅m2
    Original omega = 0.4 m/s^2
    so L = (3.4*.4) * 10^-3 = 1.36 * 10^-3 forever
    Original I about axis at near end of rod = (1/3) m * length^2 = 3.4*10^-3
    so
    m * .55^2 = 3*3.4*10^-3
    m = 33.7 * 10^-3 kg or 33.7 grams

    Final I = 3.4 *10^-3 + m * .55^3
    Final omega = v/R = 0.202 / .55 = 0.367 radians/ second
    so
    (3.4 *10^-3 + m * .55^3) * 0.367 = 1.36 * 10^-3 forever
    solve for m

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